A battery of emf `E_0=12V` is connected across a `4 m` long uniform wire having resistance `4Omega//m`. The cell of small emfs `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `2Omega` and `6Omega` respectivley are connected as shown in the figure. If galvanometer shows no diflection at the point `N` the distance of points `N` from the point `A` is equal to A. `5/3m`B. `4/3m`C. `3/2m`D. none of these

A battery of emf `E_0=12V` is connected across a `4 m` long uniform wi

1.	A battery of emf `E_0=12V` is connected across a `4 m` long uniform wire having resistance `4Omega//m`. The cell of small emfs `epsilon_1=2V` and `epsilon_2=4V` having internal resistance `2Omega` and `6Omega` respectivley are connected as shown in the figure. If galvanometer shows no diflection at the point `N` the distance of points `N` from the point `A` is equal to A. `5/3m`B. `4/3m`C. `3/2m`D. none of these
Answer» Correct Answer - D Equivalent emf of two batteires `epsilon_1` and `epsilon_2` is `epsilon=(epsilon_1//r_1+epsilon_2//r_2)/(1//r_1+1//r_2)=((2//2)+(4//6))/((1//2)+(1//6))` `2.5V` Now, `V_(AN)=epsilon` `:. I_(AN)=(R_(AN))=epsilon` or `((12)/(4+4xx4))(4)(l)=2.5` Solving this equation we get `l=2.5/24m`

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